Solution:

Without loss of generality, we may assume that $A, B, C$, and $D$ have the relative positions shown. We label the points of tangency $W, X, Y$, and $Z$. We also note that the entire construction is symmetric about the line of centers $OP$, which therefore perpendicularly bisects segments $AB$, $CD$, $WX$, and $YZ$ at their respective midpoints $K, L, M$, and $N$. Let $r_1$ and $r_2$ be the respective radii of $j$ and $k$. We will first prove that $AK = CL$. Since $\triangle OAK \sim \triangle OYN \sim \triangle OPY$, we have
$$\frac{AK}{OA} = \frac{PY}{OP}, \quad \text{so } AK = \frac{OA \cdot PY}{OP} = \frac{r_1 r_2}{OP}$$
Symmetrically, $CL = r_1 r_2 / OP$ so $AK = CL$. Now quadrilateral $AKLC$ has right angles at $K$ and $L$ and equal, parallel sides $AK = CL$, so it is a rectangle. Symmetrically, $BKLD$ is a rectangle so $ABDC$ is a rectangle.